proving with prime numbers: For which number a no number of a+2,a+3,...,a+(n+1) is...

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lostinmath

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Hello i need so help. I'm not sure if this goes under pre-algebra, i don't even know where this goes but i need some help and if i'm in the wrong section please direct me to the correct one :D

I have to prove that for every natural number n exists n-number of numbers following one another and none of this is a prime number. A bonus clue the professor gave us was: For which number a no number of a+2,a+3,...,a+(n+1) is a prime number. And i don't even know how this is supposed to be true, for example if we take a number 5 that means i have to prove that none of the 5 following numbers are prime numbers... Those numbers are 6,7,8,9,10 of which 7 is a prime number... if someone could help me i'd be really thankful! :p
 
well i missunderstood it* it means that if you pick any natural number you will find two prime numbers with the exact number of naturaln number in between.. like if i take 10000000 there are two prime numbers which have exactly 10000000 natural numbers between and there is no other prime number... how can i prove that ?
 
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Uh... what? I'm sorry, but I'm having a hard time understanding any of what you posted. It looks to me as though there's actually two different problem statements in there. The first one seems to be something like the following, but I'm really just taking a wild stab in the dark here:

Prove that, for any natural number n, no number in the set {n+1, n+2, ..., 2n-1} is prime.
Click to expand...

But, if that's the case, that's easily falsifiable, as you noted. Given n=5, the set is {6, 7, 8, 9}, of which 7 is prime. Thus, we've proven the conjecture untrue.

Your second statement, which appears to me to be a separate problem altogether, seems something like this:

Prove that, for any natural number a, no number in the set {a+2, a+3, ..., a+n+1} is prime.
Click to expand...

But, here, I'm confused because now we have two variables, but only one is explicitly declared to mean anything? What does n signify in this context? Is n also allowed to be any natural number? If that's the case, this can also easily be invalidated. Let a=n=5, and the set is {6, 7, 8, 9, 10, 11}, of which 7 and 11 are prime. Thus, this is also false.

I think it might help if you post the full and exact text of the problem as given to you by your instructor, so that we're all operating from the same information.
 
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lostinmath said:
Hello i need so help. I'm not sure if this goes under pre-algebra, i don't even know where this goes but i need some help and if i'm in the wrong section please direct me to the correct one :D

I have to prove that for every natural number n exists n-number of numbers following one another and none of this is a prime number. A bonus clue the professor gave us was: For which number a no number of a+2,a+3,...,a+(n+1) is a prime number. And i don't even know how this is supposed to be true, for example if we take a number 5 that means i have to prove that none of the 5 following numbers are prime numbers... Those numbers are 6,7,8,9,10 of which 7 is a prime number... if someone could help me i'd be really thankful! :p
Click to expand...
for example if we take a number 5 that means you must find some sequence of 5 numbers, none of which is prime. So consider any sequence of 5 numbers. They must be of the form
a+2, a+3, a+4, a+5, a+6
for some number a. Now which number a makes the above sequence divisible by 2, 3, 4, 5, and 6.
>>>>>>>>>>>>>>>>>>>>>>>>>
How about 7! Highlight, for an answer
<<<<<<<<<<<<<<<<<<<<<<<<<
 
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I made a mistake I misunderstood, my bad. N is meant to be any natural number. He wants me to prove that if I pick any natural number n there are two prime numbers with that exact number of natural number in between... for example if I take a number 100.. then two prime numbers exist which have 100 natural number between them and no other prime number. and for the clue he didn't say anything if the number a is natural or any other, but it probably is natural since he didn't say otherwise. I'm sorry I'm translating this from my mother tongue and some translations might be misleading and I'd be glad to give more explanation if needed. I'm not sure if that is your stab in the dark ksdhart 2 but I think it's pretty close :)
 
lostinmath said:
I made a mistake I misunderstood, my bad. N is meant to be any natural number. He wants me to prove that if I pick any natural number n there are two prime numbers with that exact number of natural number in between... for example if I take a number 100.. then two prime numbers exist which have 100 natural number between them and no other prime number. and for the clue he didn't say anything if the number a is natural or any other, but it probably is natural since he didn't say otherwise. I'm sorry I'm translating this from my mother tongue and some translations might be misleading and I'd be glad to give more explanation if needed. I'm not sure if that is your stab in the dark ksdhart 2 but I think it's pretty close :)
Click to expand...
The problem "Given a natural number n, there is some sequence of natural numbers of length n (or more) for which each number in the sequence is a composite number. The number preceding the sequence and the number following the sequence may or may not be prime" is a problem given very often in the study of natural numbers, i.e. Number Theory in USA English. The bonus clue given by the professor gives a as an integer function of n which provides the sequence. That is there exists a function a(n) such that all numbers in the sequence
a(n)+2, a(n)+3, a(n)+4, ... a(n)+n+1
are composite.

As an example of the proof for 3 (and 4 and 5 and 6 and 7 which is as far as I have gone): There is a sequence of numbers of length 3 for which all 3 numbers are composite, i.e. (122, 123, 124) which has composites at both ends. Note that if you add the preceding and/or following numbers to the sequence, it can serve as an example for 5 and/or 4. However, that is not how I would do the proof for the statement.
 
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Oh, okay. I think I understand now. To start with, I'd distill what you're trying to prove down to its core. If we let pn be the nth prime number, and gn be the gap between two consecutive primes, then:

gn = pn+1 - pn

For reasons which should be obvious, we already know that gn must be a natural number. So, the goal is to prove:

\(\displaystyle \displaystyle \limsup _{n\to \infty } \left(g_n\right)=\infty\)

In other words, the maximum value of gn is infinity (i.e. gn can be any natural number).

Now, to aid in the proof, you might consider the sequence of numbers {n! + 2, n! + 3, ..., n! + n} for any n > 1. What do you notice about these numbers? What are their prime factors? Given that information, suppose that n! + 1 is prime (it won't always be). What does that tell you about the gap between n! + 1 and the next prime? What if n! + 1 isn't prime? Does that change anything?
 
ksdhart2 said:
Oh, okay. I think I understand now. To start with, I'd distill what you're trying to prove down to its core. If we let pn be the nth prime number, and gn be the gap between two consecutive primes, then:

gn = pn+1 - pn

For reasons which should be obvious, we already know that gn must be a natural number. So, the goal is to prove:

\(\displaystyle \displaystyle \limsup _{n\to \infty } \left(g_n\right)=\infty\)

In other words, the maximum value of gn is infinity (i.e. gn can be any natural number).

Now, to aid in the proof, you might consider the sequence of numbers {n! + 2, n! + 3, ..., n! + n} for any n > 1. What do you notice about these numbers? What are their prime factors? Given that information, suppose that n! + 1 is prime (it won't always be). What does that tell you about the gap between n! + 1 and the next prime? What if n! + 1 isn't prime? Does that change anything?
Click to expand...
If that is the statement of the problem, rather than
\(\displaystyle \displaystyle \limsup _{n\to \infty } \left(g_n\right)=\infty\)
wouldn't the proof have to be something like "there exists a map from {g(n)| g(n)=p(n+1)-p(n); p(n) the nth prime} onto {1, 2, 3, 4, ...}. That is, not only is the \(\displaystyle \limsup \infty\) but it must 'touch' every number in between 1 and \(\displaystyle \infty\).
 
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